# Statistic HELP! Hope someone can

1. Dec 15, 2004

### johnnyICON

Statistic HELP! Hope someone can :(

I know this is only a physics and math help forum, but I'm pretty sure that most of you have taken a course in statistics... or so I hope. I'll take a long shot anyway...

I have a question where I need to find the sample size required in order to get a margin of error that is 0.9.

Here is the question:
A random sample size of 25 has been taken from a population in order to estimate the population mean. The standard deviation of the population is known. The sample mean is 10.2, and the associated margin of error, at 90% level of confidence, is equal to 1.8 If you want to reduce the margin of error to one-half of this value, using the same level of confidence, what would be the minimum sample size required?

The only formula I know in relation to margin of error is:
$$n = \frac{z^*\sigma}{m}$$

• where z* is the critical value derived the the confidence level, that being 1.645 for this matter.
• sigma is the standard deviation, which I DO NOT KNOW!
• m is the margin of error, which will be 0.9

I've had two question's like this already. My real problem is figuring out how to find sigma, the standard deviation. If anyone can help... PLEASE DO! Thanks.

2. Dec 15, 2004

### Andrew Mason

First of all, I think you have to square the right side:

(1) $$n = |\frac{z^*\sigma}{m}|^2$$

This allows you to derive $\sigma$ from the margin of error (which only makes sense, since it was used to calculate the margin of error):

(2) $$\sigma = \sqrt{n}\frac{m}{z^*}$$

Then plug it back into (1) to find n for m = .9 . But you really don't need to know $\sigma$. All you need to do is know that $n \propto \frac{1}{m^2}$

AM

Last edited: Dec 15, 2004
3. Dec 15, 2004

### johnnyICON

AWESOME MAN, THANKS.

I should of saw that... I guess all this stress and anger is blinding me!!